By Sylow's 1st theorem, G has a subgroup of order 11. let [itex]n_p[/itex] be the number of sylow p-subgroups. then [itex]n_{11} = 1(mod 11)[/itex] and [itex]n_{11}[/itex] divides 2^2 (=4) so [itex]n_{11}=1[/itex]. Therefore, it must be a normal subgroup (since it has no distinct conjugates).

Well, what are your thoughts? The obvious thing is to think about the Sylow 2s. There's either 1 or 11 of them (why?). If there's just 1, say H, then G is isomorphic to the direct product of H and the Sylow 11 (why?). Now what?

Scenario 2: there's 11 Sylow 2s. Take one of them and play around with it and with the Sylow 11. Remember that the Sylow 11 is cyclic and normal in G -- this will probably be useful in getting a presentation for G.